A few days ago
ahma914

# STATISTICS HELP on Central Limit Theorem (Statistics for Management)?

PLEASE!! i need these two questions to be solved emergency!!! thanks for everyone….

1- by reviewing sales since opening 6 months ago, a restaurant owner found that the average bill for a couple was \$26, and the standard deviation was \$5.65. how large would a sample of customers have to be for the probability to be at least 95.44 percent that the mean cost per meal for the sample would fall between \$25 and \$27?

2-The U.S. Customs Agency routinely checks all passengers arriving from foreign countries as they enter the U.S.. The department reports that the number of people per day found to be carrying contraband material as they enter the U.S. through John F. Kennedy airport in New York averages 42 and has a standard deviation of 11. What is the probability that in five days at the airport, the average number of passengers found carrying contraband will exceed 50?

Top 1 Answers
A few days ago
ahem c

Favorite Answer

===========1. Two-tailed Z test============

Z of 95.44% / 2 = Z of 47.72% = 1.9991

-1.9991 <= our Z <= 1.9991 ( you can find z value here: http://davidmlane.com/hyperstat/z_table.html ) mean = 26\$; the value can swing to the left(25\$) or the right(27\$) but shouldn't be more extreme than those two 95.44% at a time. so, we choose 27\$ as xbar and our Z = 1.9991 = (xbar - u) / (S.D. / n^0.5) n^0.5 = 1.9991*S.D / (xbar-u) = 1.9991*5.65 / (27 - 26) n^0.5 = 11.295 ; n = 127.58 or 128 so you need 128 samples. ============ 2. Central Again? ========== Assumption: pop. mean = u = 42 pop S.D. = 11 Average number = number of people in all 5 days and average it So, we have 5 random variables ( 1 per day ), therefore, n = 5 Z = (xbar - u) / ( S.D. / n^0.5 ) Z = (50 - 42) / ( 11 / 5^0.5 ) = 1.626 "exceed 50" mean that we need find the area to the right of z. Answer: P(xbar >= 50) = P(Z >= 1.626) = 5.2%

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